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Let $k$ be a field, and $Z = \textrm{Spec } k = \{0\}$. A $k$-scheme is a scheme $(X, \mathcal O_X)$ together with a morphism of schemes $j: X \rightarrow Z$. Giving such a morphism is equivalent to giving a ring homomorphism $k \rightarrow \mathcal O_X(X)$, i.e. a $k$-algebra structure on $\mathcal O_X(X)$. This then determines, by restriction, a $k$-algebra structure on $\mathcal O_X(U)$ for each $U \subseteq X$ open. If $(Y, \mathcal O_Y)$ is another $k$-scheme with morphism $i: Y \rightarrow Z$, then a morphism of $k$-schemes is a morphism of schemes $f: X \rightarrow Y$ such that $i \circ f= j$. If for $\mathcal O_Y(U) \neq 0$, and $U$ open in $Y$, we translate to the ring homomorphism on the global sections $f^{\#}(U): \mathcal O_Y(U) \rightarrow \mathcal O_X(f^{-1}U)$, then the equality $i \circ f = j$ just seems to assert that each $f^{\#}(U)$ is the identity on $k$, i.e. a $k$-algebra homomorphsm.

So, is a $k$-scheme the same thing as a scheme, with "sheaf of rings" replaced by "sheaf of $k$-algebras." And, a morphism of $k$-schemes is the same thing as a morphism between schemes, except the ring homomorphisms of $k$-algebra homomorphisms? Is there anything more to the notion of a $k$-scheme?

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Yes, with "commutative" everywhere. But this won't generalize cleanly to schemes over other bases.

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